Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin >> N;
  vector<double> head(N+1);
  double p;
  cin >> p;
  head[0] = 1-p;
  head[1] = p;
  for(int i = 1; i < N; ++i){
    cin >> p;
    for(int j = i+1; j >= 0; --j){
      head[j] *= (1-p);
      if(j > 0) head[j] += head[j-1]*p;
    }
  }
  double ret = 0;
  for(int i = (N+1)/2; i <= N; ++i) ret += head[i];
  cout << fixed << setprecision(10) << ret;
  return 0;
}